The astute will be ahead of me already on how this dents FET, but for the simple souls such as Thork, the problem FET is faced with is one of triangulation. If you know where you are, and another observer knows where they are, and they can both see the ISS at the position it's predicted to be in from their location, then simple trigonometry can give you a good estimation of the height of the ISS and how fast it's going. It doesn't take much calculation to show that this thing is higher and faster than an aircraft, and what's more, combine these observations with data from any home astronomy program (I use Redshift) and you can see it pass into the shadow of the earth at exactly the correct place for where the sun is supposed to be located relative to the earth's surface.

So stop posting and actually do one of these thought experiments then. We get people here every day saying "if you do this... this will happen" and then think that they won.

Er, have you STILL not learned to read? Did you see the bit that says that I made zetetic observations myself? The experiment has been done, idiot. If you'd like to see some maths on it:

Altitude from southern England: 40 degrees

Altitude on same pass seen from western Wales: 71 degrees.

Distance between these two observation points is roughly 325km. If the ISS position is taken as the point of a triangle and the observation sites as the other two points, that gives an angle between the two lines of sight of 69 degrees. This is therefore not far off an isoceles triangle, which means that

the

**minimum ** distance the ISS could be from the ground to create this view is somewhere around 200km - which would be if the plane of the triangle was perpendicular to a line between the two observing locations. However, it's not, it's actually leaning over somewhat, which means the triangle plane described between these three points is "stretched out" in the direction of the ISS. I'm not sure what mathematics is required in order to account for this, presumably by adding in a third observation point, say from mid Scotland at 25 degrees. Then you can create a tetrahedron with only one possible vertex position, but I'm not sure exactly how that is calculated. However, even my simple working out has given a minimum height above the ground well above that of aircraft.

If anyone can do more precise calculations, please feel free to correct me.